1
$\begingroup$

We have $N\geq 9$ distinct positive reals numbers in range $[0,1)$ such that whenever we pick $8$ of them, there is another one (different from the first $8$) so that their sum is an integer. What are the possible values of $N$?

Clearly $N=9$ works, just take $9$ distinct reals adding to $1$. But I have not been able to figure out other values. However, we can notice that there are at least $\frac{\binom{N}{8}}{9}$ subsets of $9$ elements which have an integer sum, and this sum must be among $\{1,2,3,4,5,6,7,8\}$. So at least one of the numbers appears at least $\frac{\binom{N}{8}}{72}$ times. I get a feeling that this is impossible for large $N$.

$\endgroup$
0
2
$\begingroup$

Partial result: The case $N=10$ is impossible.

Pf: Suppose we had $10$ of them, sorted as $0<a_1<\cdots <a_{10}$.

let $x=a_1+\cdots a_8$ We know that either $x+a_9$ or $x+a_{10}$ is an integer.

Remark: they can't both be integers as $0<a_{10}-a_9<1$

Case I: $x+a_9$ is an integer.

Then define $y=a_2+\cdots +a_8+a_{10}$ Which of $a_1$ and $a_9$ should we add to get an integer? It can't be $a_1$ by the remark above, so it must be $a_9$. But then $(y+a_9)-(x+a_9)=y-x=a_{10}-a_1 \in \mathbb Z$, a contradiction.

Case II: $x+a_{10}$ is an integer.

Then define $y=a_2+\cdots a_{9}$ Which of $a_1$ and $a_{10}$ should we add to get an integer? It can't be $a_1$ by the remark above, so it must be $a_{10}$. But then $(y+a_{10})-(x+a_{10})=y-x=a_{9}-a_1 \in \mathbb Z$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.